viele krasse sachen zur VL1
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src/main.typ
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src/main.typ
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@ -1,24 +1,30 @@
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#import "@preview/charged-ieee:0.1.0": ieee
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#let title = [ Multivariable Analysis an der DHBW weil ich durchgefallen bin (Scheiße) ]
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#let abstract = [
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Notizen zu Vorlesungen und Lösungen von Aufgaben bezüglich multivariabler
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Analysis and der DHBW Mannheim.
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]
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#show: ieee.with(
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title: [Multivariable Analysis an der DHBW weil ich durchgefallen bin (Scheiße)],
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#import "@preview/arkheion:0.1.0": arkheion, arkheion-appendices
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#show: arkheion.with(
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title: title,
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authors: (
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(
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name: "Christoph J. Scherr",
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department: [TINF22CS2 (Cybersecurity at DHBW)],
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organization: [NewTec GmbH],
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location: [Mannheim, Germany],
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email: "contact@cscherr,de"
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(name: "Christoph J. Scherr", email: "contact@cscherr.de", affiliation: "NewTec GmbH"),
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),
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),
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index-terms: ("Math", "Undergrad"),
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bibliography: bibliography("refs.bib"),
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// Insert your abstract after the colon, wrapped in brackets.
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// Example: `abstract: [This is my abstract...]`
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abstract: abstract,
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keywords: ("Undergrad", "Analysis", "Mathmatics"),
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date: datetime.today().display(),
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)
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#set text(lang: "de")
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// Justified paragraphs
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#set par(justify: true, first-line-indent: 0pt)
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#set par(
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justify: true,
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leading: 0.52em,
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)
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// more space between pars
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#show par: set block(spacing: 2em)
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@ -26,10 +32,22 @@
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// Put this here to avoid affecting the title
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#show link: underline
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// Preamble
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// headcolor
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#let headcolor = rgb("80b3ff")
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#set figure(numbering:"1.1")
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// Preabmle
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///////////////////////////////////////////////////////////////////////////////////////////////////
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#outline()
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#v(8em)
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#outline(depth: 3)
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#outline(title: "Abbildungen", target: figure)
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#pagebreak()
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// Content
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///////////////////////////////////////////////////////////////////////////////////////////////////
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@ -41,17 +59,26 @@ brauchte einfach mehr Zeit. Dieses Dokument wird meine Notizen und Lösungen
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zu allen Vorlesungen und Übungen enthalten.
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Dieses Dokument bezieht sich vor allem auf Vorlesung#cite(<Vorlesung>) und Übungen#cite(<Exercise>).
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Diese sind auf Englisch verfasst, dieses Dokument wird jedoch auf Deutsch sein.
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Diese sind auf Englisch
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= Methoden
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Erstmal machen und gucken dann.
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#pagebreak()
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= Vorlesungen
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#include "./vorlesungen/index.typ"
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#pagebreak()
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= Übungen
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#include "./exercise/index.typ"
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// Postabmle
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///////////////////////////////////////////////////////////////////////////////////////////////////
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#pagebreak()
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#bibliography("refs.bib")
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@ -9,3 +9,10 @@
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title = "Applied Mathematics - Multivariable analysis (Übungen)",
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year = "2024",
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}
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@online{MatheNichtFreaks,
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url = "
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https://de.wikibooks.org/wiki/Mathe_f%C3%BCr_Nicht-Freaks:_Teleskopsumme_und_Teleskopreihe
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",
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urldate = "2024-07-18",
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}
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@ -0,0 +1,214 @@
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== Basics & Outlook @Vorlesung[Foliensatz 1]
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- Größtenteils Wiederholung von Analysis
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=== Konvergenz von Folgen
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#v(2em)
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#figure([
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+ Ist die Folge von einem bestimmten Typen?
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- Harmonische Reihe ($1/n$)
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- Alternierende Reihe ($(-1)^n$)
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+ Hilft die 3. Bionische Formel? ($(a+b)(a-b)$) (heiß´t glaub ich auch
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quadratische Ergänzung)
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+ Sandwich Theorem ($a_n <= b_n <= c_n => a <= b <= c$)
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+ Gibt es einen Maximalwert?
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+ Ist sie monoton: ($a_(n+1)/a_n <= 1 "oder" >= 1$)
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], caption: "Algorithmus zum Test auf Konvergenz bei Folgen") <alg:folge>
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==== Ex: Convergence of Sequences @Vorlesung[1, P. 11]
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#quote()[
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Does the sequence $a_n; n in NN$
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$ a_n = ((n+1)^2 + 3 dot n(n^2 -1)) / (2 dot (n+2)^3) $
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converge? If it converges, what is the limit?
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]
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$
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a_n &= ((n+1)^2 + 3 dot n(n^2 -1)) / (2 dot (n+2)^3)) \
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=> lim_(n -> infinity) a_n &= lim_(n -> infinity) ((n+1)^2 + 3 dot n(n^2 -1)) / (2 dot (n+2)^3) \
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&= lim_(n -> infinity) (3 n^3) / (2n^3) \
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&= 3 / 2 checkmark
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$
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==== Ex: Convergence of Sequences @Vorlesung[1, P. 13]
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#quote()[
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Does the sequence $a_n; n in NN$
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$ a_n = sqrt(n^2+n)-n $
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converge? If it converges, what is the limit?
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]
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$
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a_n &= sqrt(n^2+n)-n \
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=> lim_(n -> infinity) a_n &= lim_(n -> infinity) sqrt(n^2+n)-n \
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&= lim_(n -> infinity) sqrt(n^2+n)-n \
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&> lim_(n -> infinity) sqrt(n^2)-n = 0 \
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lim_(n -> infinity) a_n &= lim_(n -> infinity) sqrt(n^2+n)-n \
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&= lim_(n -> infinity) (sqrt(n^2+n)-n) dot (sqrt(n^2+n)+n)/(sqrt(n^2+n)+n) \
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&= lim_(n -> infinity) ((sqrt(n^2+n)-n) dot (sqrt(n^2+n)+n))/(sqrt(n^2+n)+n) \
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&= lim_(n -> infinity) (n^2+n-n^2)/(sqrt(n^2+n)+n) \
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&= lim_(n -> infinity) (n)/(sqrt(n^2+n)+n) |:n\
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&= lim_(n -> infinity) (1)/(sqrt(n^2+n)/n +1)\
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&= lim_(n -> infinity) (1)/(sqrt(n^2+n)/sqrt(n^2) +1)\
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&= lim_(n -> infinity) (1)/(sqrt((n^2+n)/n^2) +1)\
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&= lim_(n -> infinity) (1)/(sqrt(1+1/n) +1) = 1/2 checkmark
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$
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==== Ex: Convergence of Sequences @Vorlesung[1, P. 15]
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#quote()[
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Does the sequence $a_n; n in NN$
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$ a_n = n!/n^n $
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converge? If it converges, what is the limit?
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]
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$
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a_n &= n!/n^n \
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=> lim_(n -> infinity) a_n &= lim_(n -> infinity) n!/n^n \
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&= ((n-1)(n-2)dots)/n^n \
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=> n! &<= n^(n-1) \
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=> a_n &<= n^(n-1)/n^n = 1/n --> 0 \
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=> lim_(n -> infinity) a_n &= 0
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$
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=== Konvergenz von Reihen
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#v(2em)
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#figure([
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+ Addiert die Reihe eine Nullfolge?
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+ Ist die Reihe von einem bestimmten Typen?
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- Geometrische Reihe
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- Teleskop-Reihe
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+ Addiert die Reihe eine alternierende Folge?
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+ Ratio test (Besonders bei $x!$ und $x^n$)
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+ Root test (Besonders bei $x^n$)
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], caption: "Algorithmus zum Test auf Konvergenz bei Reihen") <alg:reihe>
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==== Ex: Convergence of Serieses @Vorlesung[1, P. 22]
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#quote()[
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Does the series
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$ S = sum^(infinity)_(n=0) (2n-1)/(3n-1) $
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converge?
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]
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$
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lim_(n -> infinity) (2n-1)/(3n-1) = 2/3 != 0 \
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=> "Die Reihe konvergiert nicht, weil " (2n-1)/(3n-1) " keine Nullfolge ist." checkmark
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$
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#pagebreak()
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==== Ex: Convergence of Serieses @Vorlesung[1, P. 26]
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#quote()[
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Does the series
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$ S = sum^(infinity)_(n=0) (2)/(3^n) $
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converge?
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]
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$
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S &= sum^(infinity)_(n=0) (2)/(3^n) \
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&= 2 dot sum^(infinity)_(n=0) (1)/(3^n) \
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&= 2 dot sum^(infinity)_(n=0) (1/3)^n ; 1/3 > 1 \
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&=> S " ist eine geometrische Reihe!" \
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S &= 2 dot 1/(1- 1/3) = 2 dot 1/(2/3) = 2 dot 3/2 = 3 checkmark
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$
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==== Ex: Convergence of Serieses @Vorlesung[1, P. 29]
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#quote()[
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Does the series
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$ S = sum^(infinity)_(n=0) (1)/(n(n+1)) $
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converge?
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]
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$
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S &= sum^(infinity)_(n=0) (1)/(n(n+1)) \
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b_n &= (1)/(n(n+1)) \
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&= (1+(n-n))/(n(n+1)) \
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&= ((1+n)-n)/(n(n+1)) \
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&= ((n+1))/(n(n+1)) - (n)/(n(n+1)) \
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&= (1)/(n) - (1)/(n+1) \
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"mit " a_n = 1/n &: b_n = a_n - a_(n+1) => "Wir können das Teleskopkriterium anwenden" \
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(S &= a_0 = 1/1 = 1 checkmark) \
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S &= a_1 = 1/1 = 1 checkmark \
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$
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Diese Partialbruchzerlegung hätte ich nicht ohne
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@MatheNichtFreaks[Partialbruchzerlegung] gefunden. Aus irgendeinem Grund
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ist $a_0 = 1/1$ in der Vorlesung@Vorlesung, anstatt $1/0$. Demnach müsste
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z.B. $a_1=1/2$ sein. Das $n$ in $a_n$ indiziert also eine Zahl aus $NN$
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und wird nicht direkt eingesetzt.
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Ich finde das ist inkonsistent, deshalb verwende ich $a_1 => n=1$, und $a_0 => n=0$.
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#pagebreak()
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==== Ex: Convergence of Serieses @Vorlesung[1, P. 32]
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#quote()[
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Does the series
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$ S = sum^(infinity)_(n=0) (n)/(2^n) $
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converge?
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]
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$
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lim_(n->infinity) abs((a_(n+1)/a_n)) &= lim_(n->infinity) abs(((n+1)/(2^(n+1)))/(n/(2^n))) \
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&= lim_(n->infinity) abs(((n+1) dot 2^n)/(n dot 2^(n+1))) \
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&= lim_(n->infinity) abs(((n+1))/n) dot abs((2^n)/(2^(n+1))) \
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&= lim_(n->infinity) abs(((n+1))/n) dot 1/2 \
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&= lim_(n->infinity) abs((n)/n) dot 1/2 \
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&= 1/2 checkmark
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$
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==== Ex: Convergence of Serieses @Vorlesung[1, P. 35]
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#quote()[
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Does the series
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$ S = sum^(infinity)_(n=0) (2^n)/((n+1)^n) $
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converge?
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]
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$
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&lim_(n->infinity) root(n, abs((2^n)/((n+1)^n))) "(Wurzelkriterium)"\
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= &lim_(n->infinity) root(n, abs(((2)/((n+1)))^n)) \
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= &lim_(n->infinity) (2)/((n+1)) = 0 < 1 => "S ist konvergent" checkmark
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$
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=== Funktionen
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==== Ex: Convergence of Serieses @Vorlesung[1, P. 35]
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#quote()[
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Does the series
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$ S = sum^(infinity)_(n=0) (2^n)/((n+1)^n) $
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converge?
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]
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$
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&lim_(n->infinity) root(n, abs((2^n)/((n+1)^n))) "(Wurzelkriterium)"\
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= &lim_(n->infinity) root(n, abs(((2)/((n+1)))^n)) \
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= &lim_(n->infinity) (2)/((n+1)) = 0 < 1 => "S ist konvergent" checkmark
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$
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@ -1,20 +1 @@
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== Basics & Outlook @Vorlesung[Foliensatz 1]
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- Größtenteils Wiederholung von Analysis
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=== Ex: Convergence @Vorlesung[P. 11]
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#quote(attribution: [@Vorlesung[P. 11]])[
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Does the sequence $a_n; n in NN$
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$ a_n = ((n+1)^2 + 3 dot n(n^2 -1)) / (2 dot (n+2)^3) $
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converge? If it converges, what is the limit?
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]
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$
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a_n &= ((n+1)^2 + 3 dot n(n^2 -1)) / (2 dot (n+2)^3)) \
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=> lim_(n -> infinity) a_n &= lim_(n -> infinity) ((n+1)^2 + 3 dot n(n^2 -1)) / (2 dot (n+2)^3) \
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&= lim_(n -> infinity) (3 n^3) / (2n^3) \
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&= 3 / 2 checkmark
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$
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#include "1.typ"
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